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If a Hausdorff space $\ X\ $ admits a dense subset $ A \hookrightarrow X\ $ such that

$$|X|^{|A|}\ =\ |X|$$

then indeed $$|X| \leq |\text{End}_{\text{Top}}(X)| \leq |X|^{|A|}\ = \ |X|.$$

It is the case of $\mathbb{Q} \hookrightarrow \mathbb{R}$. Thus, if there is a small enough dense subspace, there are not so many endomorphisms.

Is the converse true?

Namely, suppose $|\text{End}_{\text{Top}}(X)| = \ |X|.$ Is it true that there exists a dense subset $A$ such that $|X|^{|A|}\ =\ |X|$?


UPDATE: Solved on MO.

1 Answers1

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(To kick it from the unanswered queue)

As mentioned by the OP, this is solved on MO: Some T2 spaces must have a small dense?


@KeithKearnes 's solution:

Is the converse true?

No.

The paper Constructions and Applications of Rigid Spaces, I, Advances in Mathematics 29 (1978), 89-130, by Kannan and Rajagopalan, describes a countaby infinite Hausdorff space $X$ such that the only continuous maps $f\colon X\to X$ are the constant maps and the identity map. Such a space satisfies $|\text{End}_{\text{Top}}(X)| = \omega = \ |X|.$

If $A\subseteq X$ is a subset such that $|X|^{|A|}=|X|=\omega$, then $A$ is forced to be finite, and therefore $A$ cannot be dense.

YuiTo Cheng
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